Let `lambda, mu in R`. If the system of equations
`3x + 5y + lambda z = 3`
`7x + 11y - 9z = 2`
`97x + 155y - 189z = mu`
has infinitely many solutions, then `mu + 2 lambda` is equal to:

Options

A. 24
B. 27
C. 25
D. 22

Answer

To solve the problem, we need to find the value of \( \mu + 2\lambda \) given the system of equations has infinitely many solutions. The equations are:

1. \( 3x + 5y + \lambda z = 3 \)
2. \( 7x + 11y - 9z = 2 \)
3. \( 97x + 155y - 189z = \mu \)

### Step 1: Set up the determinant condition

For the system of equations to have infinitely many solutions, the determinant of the coefficient matrix must be zero. The coefficient matrix \( A \) is:

\[ A = \begin{bmatrix} 3 & 5 & \lambda \\ 7 & 11 & -9 \\ 97 & 155 & -189 \end{bmatrix} \]

We need to calculate the determinant \( \Delta = |A| \) and set it equal to zero.

### Step 2: Calculate the determinant

Using the formula for the determinant of a 3x3 matrix:

\[ |A| = a(ei - fh) - b(di - fg) + c(dh - eg) \]

where the matrix is represented as:

\[ \begin{bmatrix} a & b & c \\ d & e & f \\ g & h & i \end{bmatrix} \]

We can expand the determinant along the first row:

\[ \Delta = 3 \begin{vmatrix} 11 & -9 \\ 155 & -189 \end{vmatrix} - 5 \begin{vmatrix} 7 & -9 \\ 97 & -189 \end{vmatrix} + \lambda \begin{vmatrix} 7 & 11 \\ 97 & 155 \end{vmatrix} \]

Calculating each of these 2x2 determinants:

1. \( \begin{vmatrix} 11 & -9 \\ 155 & -189 \end{vmatrix} = (11)(-189) - (-9)(155) = -2079 + 1395 = -684 \)
2. \( \begin{vmatrix} 7 & -9 \\ 97 & -189 \end{vmatrix} = (7)(-189) - (-9)(97) = -1323 + 873 = -450 \)
3. \( \begin{vmatrix} 7 & 11 \\ 97 & 155 \end{vmatrix} = (7)(155) - (11)(97) = 1085 - 1067 = 18 \)

Substituting these values back into the determinant:

\[ \Delta = 3(-684) - 5(-450) + \lambda(18) \]
\[ = -2052 + 2250 + 18\lambda \]
\[ = 198 + 18\lambda \]

Setting the determinant equal to zero for infinitely many solutions:

\[ 198 + 18\lambda = 0 \]

### Step 3: Solve for \( \lambda \)

\[ 18\lambda = -198 \]
\[ \lambda = -11 \]

### Step 4: Find \( \mu \)

Next, we need to find \( \mu \) using the condition that the ratios of the determinants must also be equal. We can use the determinant of the matrix formed by replacing the third column with the constants from the equations:

\[ \Delta_z = \begin{vmatrix} 3 & 5 & 3 \\ 7 & 11 & 2 \\ 97 & 155 & \mu \end{vmatrix} \]

Calculating this determinant similarly:

\[ \Delta_z = 3 \begin{vmatrix} 11 & 2 \\ 155 & \mu \end{vmatrix} - 5 \begin{vmatrix} 7 & 2 \\ 97 & \mu \end{vmatrix} + 3 \begin{vmatrix} 7 & 11 \\ 97 & 155 \end{vmatrix} \]

Calculating the 2x2 determinants:

1. \( \begin{vmatrix} 11 & 2 \\ 155 & \mu \end{vmatrix} = 11\mu - 310 \)
2. \( \begin{vmatrix} 7 & 2 \\ 97 & \mu \end{vmatrix} = 7\mu - 194 \)
3. \( \begin{vmatrix} 7 & 11 \\ 97 & 155 \end{vmatrix} = 18 \) (as calculated before)

Substituting back into \( \Delta_z \):

\[ \Delta_z = 3(11\mu - 310) - 5(7\mu - 194) + 3(18) \]
\[ = 33\mu - 930 - 35\mu + 970 + 54 \]
\[ = -2\mu + 94 \]

Setting this equal to zero gives:

\[ -2\mu + 94 = 0 \]
\[ 2\mu = 94 \]
\[ \mu = 47 \]

### Step 5: Calculate \( \mu + 2\lambda \)

Now we can find \( \mu + 2\lambda \):

\[ \mu + 2\lambda = 47 + 2(-11) = 47 - 22 = 25 \]

Thus, the final answer is:

\[ \boxed{25} \]